9.2 Capacitance of a Capacitor Partially Filled with Dielectric

3.1 When equal and opposite charges placed on plates
3.2 When unequal charges are placed on the two plates

6.1 Working method to find capacitance of any capacitor

7.1 Force between the Plates of a Capacitor
7.2 Energy stored in a charged conductor

8.1 Loss of energy during redistribution of charge

9.1 Effect of Dielectric
9.2 Capacitance of a Capacitor Partially Filled with Dielectric
9.3 Quantities after inserting dielectric in a capacitor (fully)

12.1 Important points in $R$-$C$ circuits

15.1 Principle of Operation
15.2 Construction

Let, a dielectric is partially filled with a dielectric of dielectric constant =$K$ as shown in figure.

If a charge $q$ is given to the capacitor, a charge $q_i$ induces on the dielectric.

where, ${q_i} = q\left( {1 - \frac{1}{K}} \right)$

Now, assume that the electric field in the region in which the dielectric is absent is ${E_ \circ }$ and where the dielectric is present is $E = {E_ \circ }/K$. The potential difference between the plates of the capacitor is,$$\begin{equation} \begin{aligned} V = {V_ + } - {V_ - } = Et + {E_ \circ }(d - t) \\ = \frac{{{E_ \circ }}}{K}t = {E_ \circ }(d - t) = {E_ \circ }\left( {d - t + \frac{t}{K}} \right) \\ = \frac{q}{{A{\varepsilon _ \circ }}}\left( {d - t + \frac{t}{K}} \right) \\\end{aligned} \end{equation} $$

Now, as per the definition of capacitance,$$C = \frac{q}{V} = \frac{{{\varepsilon _ \circ }A}}{{d - t + \frac{t}{K}}}$$

or $$C = \frac{{{\varepsilon _ \circ }A}}{{d - t + \frac{t}{K}}}$$